Question

What is the axis of symmetry of the function P(n)

P(n) = -250n^2 + 3,250n - 9,000

Answer 1

@ninjasandtigers

Answer 2

If you can find the vertex of the parabola, you have done 99.9% of the work.

Answer 3

Do you know how to do that?

Answer 4

nope

Answer 5

What do you know about parabolas?

Answer 6

Do you know the vertex form of a parabola?

Answer 7

ya

Answer 8

y=a(x-h)^2+k

Answer 9

If I give you the function f(x)=a(x-h)^2+k, would you know how to determine the vertex?
This is called the vertex form because it gives you the vertex.
What is the vertex for f(x)=a(x-h)^2+k?

Answer 10

i have no clue how to do that

Answer 11

We can use the things we learned about transformations and the graph of y=x^2 to determine that the vertex of f(x)=a(x-h)^2+k is (h,k)

Answer 12

4 and 9??

Answer 13

how did you get that?

Answer 14

i completed the square

Answer 15

we need to write P(n)=-250n^2 + 3,250n - 9,000
in the following form P(n)=a(n-h)^2+k where (h,k) will be the vertex.
I will give you the recipe to follow.

First
pretend we have P(n)=ax^2+bx+c
the first thing we want to do is factor a from the first two terms
the first terms being ax^2 and bx

Now I know you don't see in a in bx but you could write bx as abx/a
since a/a is 1 and abx/a is still bx

so anyways
P(n)=ax^2+abx/a+c

Factor out out a from the first two terms like so
P(n)=a(x^2+bx/a)+c <--do this as a first step with your function
then I will give the next step

Answer 16

whattt im so confused i dont even know what this is

Answer 17

what is?

Answer 18

oops I put x's instead of n's

P(n)=an^2+abn/a+c
P(n)=a(n^2+bn/a)+c

Answer 19

All this is saying I want you to look at the coeifficent of n^2
which your a here is -250
and factor out your a from the first two terms
that is all the first step requires

Answer 20

vertex= 6.5 and 1562.5

Answer 21

i found the vertex

Answer 22

well the vertex would be in this form (h,k)
I guess you are saying h is 6.5 and k is 1562.5

bascially the vertex is the center of the graph
or what i mean is it is where you can draw a line and see symmetry
I will give a rough picture of it

You are suppose to be able to fold your graph about that vertical line I drew and it is suppose to lay perfectly on the other side.
Do you see how that vertical line goes through the vertex?
If you know the x-coordinate of the vertex and you know that vertical lines are x=constant then you should be able to come up with the axis of symmetry (the line that divides the graph into two equal halves)